So you have a continuous-time function $x(t)$ with continuous-time Fourier Transform
$$X(j \Omega)\triangleq \mathscr{F} \Big\{ x(t) \Big\} = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j \Omega t} \, dt$$
and you have the corresponding discrete-time function $\tilde{x}[n]$ and Discrete-Time Fourier Transform
$$\tilde{X}(e^{j \omega}) \triangleq \mathcal{DTFT} \Big\{ \tilde{x}[n] \Big\} = \sum\limits_{n=-\infty}^{\infty} \tilde{x}[n] \, e^{-j \omega n}$$
The argument to $\tilde{X}(e^{j \omega})$ is periodic in $\omega$ with period of $2 \pi$.
$$ e^{j (\omega + 2 \pi)} = e^{j \omega}\,e^{j2\pi} = e^{j \omega} \cdot 1 = e^{j \omega} \qquad \forall \omega \in \mathbb{R} $$
This means
$$ \tilde{X}(e^{j (\omega + 2 \pi)}) = \tilde{X}(e^{j \omega}) \qquad \forall \omega \in \mathbb{R} $$
This is why I denote frequency-domain spectrum $\tilde{X}(e^{j \omega})$ and it's inverse DTFT, $\tilde{x}[n]$ with a little wavy tilde to denote that we know $\tilde{X}(e^{j \omega})$ is periodic.
You're sampling continuous-time function $x(t)$ at a sample rate of $f_\text{s} \triangleq \frac{1}{T}$ ($T>0$ is the sampling period). We say that the samples are
$$ \tilde{x}[n] \triangleq x(nT) \qquad n \in \mathbb{Z}$$
but that is not enough to describe the situation. Using a scaling convention that is a little different than most textbooks, the continuous-time representation of the sampled signal is:
$$\begin{align}
x_\text{s}(t) &= x(t) \left( T \sum\limits_{n=-\infty}^{\infty} \delta(t-nT) \right) \\
&= T \sum\limits_{n=-\infty}^{\infty} x(t) \delta(t-nT) \\
&= T \sum\limits_{n=-\infty}^{\infty} x(nT) \delta(t-nT) \\
&= T \sum\limits_{n=-\infty}^{\infty} \tilde{x}[n] \delta(t-nT) \\
\end{align}$$
$x_\text{s}(t)$ is a continuous-time signal, but is zero for all time, $t$, except when $t=nT$, when it is a dirac impulse function. This throws away all information about $x(t)$ between these sampling instances when $t=nT$, which is essentially what uniform sampling is all about.
So what the Poisson Summation Formula and the sampling theorem tell us is that when we uniformly sample in one domain (say the "time domain"), this causes a periodic extension to happen in the reciprocal domain (here the "frequency domain"):
$$\begin{align}
X_\text{s}(j\Omega) &= \mathscr{F} \Big\{ x_\text{s}(t) \Big\} \\
&= \mathscr{F} \Bigg\{x(t) \left( T \sum\limits_{n=-\infty}^{\infty} \delta(t-nT) \right) \Bigg\} \\
&= \mathscr{F} \Big\{T \sum\limits_{n=-\infty}^{\infty} \tilde{x}[n] \delta(t-nT) \Big\} \\
&= T \sum\limits_{n=-\infty}^{\infty} \tilde{x}[n] \mathscr{F}\Big\{ \delta(t-nT) \Big\} \\
&= T \sum\limits_{n=-\infty}^{\infty} \tilde{x}[n] e^{-j \Omega \, nT} \\
&= T \sum\limits_{n=-\infty}^{\infty} \tilde{x}[n] e^{-j \omega n} \Bigg|_{\omega = \Omega T} \\
\end{align}$$
which is $T$ times the DTFT
$$ X_\text{s}(j\Omega) = T \cdot \tilde{X}(e^{j \omega}) \Bigg|_{\omega = \Omega T} $$
The reason I include this $T$ factor to scale is for dimensional clarity. Because $x(t)$ and $\tilde{x}[n]$ have the same dimension, then in the integral of the Fourier Transform, $X(j\Omega)$ (and later $X_\text{s}(j\Omega)$) pick up an addition dimensional factor of time. Having the $T$ on the right is necessary for both sides of the equation to be dimensionally equivalent.
Another way to look at it is to consider that the sampling operator is periodic, so it has a Fourier series
$$ T \sum\limits_{n=-\infty}^{\infty} \delta(t-nT) = \sum\limits_{k=-\infty}^{\infty} e^{j 2 \pi k f_\text{s} t} $$
and all the Fourier series coefficients = 1 (only if the $T$ factor is included on the left side of the equation).
If you use the frequency translation theorem of the Fourier Transform, the continuous spectrum of that ideally sampled function is
$$\begin{align}
X_\text{s}(j\Omega) &= \mathscr{F} \Big\{ x_\text{s}(t) \Big\} \\
&= \mathscr{F} \Big\{x(t) \left( T \sum\limits_{n=-\infty}^{\infty} \delta(t-nT) \right) \Big\} \\
&= \mathscr{F} \Big\{x(t) \sum\limits_{k=-\infty}^{\infty} e^{j 2 \pi k f_\text{s} t} \Big\} \\
&= \mathscr{F} \Big\{\sum\limits_{k=-\infty}^{\infty} x(t) e^{j 2 \pi k f_\text{s} t} \Big\} \\
&= \sum\limits_{k=-\infty}^{\infty} \mathscr{F} \Big\{ x(t) e^{j 2 \pi k f_\text{s} t} \Big\} \\
&= \sum\limits_{k=-\infty}^{\infty} X\big(j (\Omega - 2 \pi k f_\text{s}) \big) \\
\end{align}$$
You can see that both spectra of the sampled signal, $x_\text{s}(t)$ and of the samples, $\tilde{x}[n]$ are periodic with period in $\Omega$ of $2 \pi f_\text{s}$ and a period in $\omega$ of $2 \pi$. The reason $f_\text{s}$ appears in the continuous-time spectrum is because the sampling frequency needs to be known for $X_\text{s}(j\Omega)$ but that and sampling period $T$ are normalized out of the expressions for both $\tilde{x}[n]$ and $\tilde{X}(e^{j \omega})$.
Equating the two expressions for the continuous-time sampled signal, $X_\text{s}(j\Omega)$:
$$ \sum\limits_{k=-\infty}^{\infty} X\big(j (\Omega - 2 \pi k f_\text{s}) \big) = T \cdot \tilde{X}(e^{j \omega}) \Bigg|_{\omega = \Omega T} $$
Now here is the last thing. If you bandlimit the continuous-time input, $x(t)$, so that
$$ X(j\Omega) = 0 \qquad \forall \ |\Omega| \ge \pi f_\text{s} $$
which means there is no energy in $x(t)$ for frequencies equal to or greater than the Nyquist frequency which is, in angular frequency, $\pi f_\text{s}$, then you know that adjacent images of $X(j\Omega)$, which are
$$ X\big(j (\Omega - 2 \pi k f_\text{s}) \big) \qquad \text{for } k \in \mathbb{Z} \ , $$
then these images do not overlap. That means when looking at the original baseband spectrum (when $k=0$)
$$ \begin{align}
X(j \Omega) &= \sum\limits_{k=-\infty}^{\infty} X\big(j (\Omega - 2 \pi k f_\text{s}) \big) \qquad & |\Omega| < \pi f_\text{s} \\
&= X_\text{s}(j \Omega) \qquad & |\Omega| < \pi f_\text{s} \\
&= T \cdot \tilde{X}(e^{j \omega}) \Bigg|_{\omega = \Omega T} \qquad & |\Omega| < \pi f_\text{s} \\
\end{align} $$
So this relates the DTFT spectrum directly to the spectrum of the original continuous-time (and bandlimited) signal.
Just for your information, the reconstruction part of the sampling theorem is that, since the samples $\tilde{x}[n]$ are sufficient information to represent $x(t)$ (this is because you bandlimited $X_\text{s}(j \Omega)$ to be non-zero only for $\Omega < 2 \pi f_\text{s}$ ), then the original $x(t)$ in terms of the samples is:
$$ x(t) = \sum\limits_{n=-\infty}^{\infty} \tilde{x}[n] \operatorname{sinc}\big( \tfrac{1}{T}(t-nT) \big) $$
where $ \operatorname{sinc}(u) \triangleq \frac{\sin(\pi u)}{\pi u} $. This is consistent with above, $x(nT)=\tilde{x}[n]$ for all integer $n$.
This explicitly relates, in both directions, the continuous-time signal $x(t)$ to the samples $\tilde{x}[n]$ and their respective spectra, $X(j \Omega)$ and $\tilde{X}(e^{j \omega})$.
